Question

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.0 kg. The cylinder pivots on a frictionless axle throughits center. The bucket is relased from rest at the top of awell and falls 10 m to the water.

a) What is the tension in the rope while the bucket isfalling?
b) what speed does the bucket strike the water?
c) What is the time of fall?
d) While the bucket is fallin, what is the force exterded onthe cylinder by the axle?

Answer 1

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

Answer 2

We have that for the Question "A 15 kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg"it can be said that

• T= 42.0 N
• v= 11.8 m/s
• t=1.69s
• N = 159.6 N

From the question we are told

A 15 kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a friction less axle through its center. The bucket is released from rest at the top of a well and falls 10 m to the water.

a) What is the tension in the rope while the bucket is falling?

b) what speed does the bucket strike the water?

c) What is the time of fall?

d) While the bucket is falling, what is the force extended on the cylinder by the axle?

a

Generally the equation for tension  is mathematically given as

$T =\frac{ Ma}{2} \\\\T =\frac{15*7}{2}$

T= 42.0 N

b)

v = \sqrt{2ah}

v = \sqrt{2*7*10}

v= 11.8 m/s

c)

v = at

Therefore

$t=\frac{v}{a}\\\\t=\frac{11.8}{7}\\\\t=1.69s$

d)

$N = mg + T\\\\Therefore\\\\N = 15*9.81 + 42.0$

N = 159.6 N

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